Find the position of the shear centre of the rectangular four boom beam section shown in Fig. P.20.4. The booms carry only direct stresses but the skin is fully effective in carrying both shear and direct stress. The area of each boom is 100 mm².

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Accepted Answer

The x axis is an axis of symmetry so that [latex]I_{xy}[/latex] = 0, also the shear centre, S, lies on this axis. Apply an arbitrary shear load, [latex]S_y[/latex], through S. The internal shear flow distribution is then given by Eq. (20.11) in which [latex]S_x[/latex] = 0 and [latex]I_{xy}[/latex] = 0. Thus

[latex]\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)+q_{s, 0}\end{aligned}[/latex] (20.11)

[latex]q_s=-\frac{S_y}{I_{x x}}\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)+q_{s, 0}[/latex] (i)

in which from Fig. S.20.4

[latex]I_{x x}=4 imes 100 imes 40^2+2 imes 0.64 imes 240 imes 40^2+\frac{0.36 imes 80^3}{12}+\frac{0.64 imes 80^3}{12}[/latex]

i.e.
[latex]I_{x x}=1.17 imes 10^6 \mathrm{~mm}^4[/latex]

‘Cut’ the section at O. Then, from the first two terms on the right-hand side of Eq. (i)

[latex]q_{\mathrm{b}, \mathrm{O} 1}=-\frac{S_y}{I_{x x}} \int_0^{s_1} 0.64 s_1 \mathrm{~d} s_1[/latex]

i.e.
[latex]q_{\mathrm{b}, \mathrm{O} 1}=-0.27 imes 10^{-6} S_y s_1^2[/latex] (ii)
and
[latex]q_{\mathrm{b}, 1}=-4.32 imes 10^{-4} S_y[/latex]
Also

[latex]q_{\mathrm{b}, 12}=-\frac{S_y}{I_{x x}}\left(\int_0^{s_2} 0.64 imes 40 \mathrm{~d} s_2+100 imes 40 ight)-4.32 imes 10^{-4} S_y[/latex]

i.e.
[latex]q_{\mathrm{b}, 12}=-10^{-4} S_y\left(0.22 s_2+38.52 ight)[/latex] (iii)
whence
[latex]q_{\mathrm{b}, 2}=-91.32 imes 10^{-4} S_y[/latex]
Finally

[latex]q_{\mathrm{b}, 23}=-\frac{S_y}{I_{x x}}\left[\int_0^{s_3} 0.36\left(40-s_3 ight) \mathrm{d} s_3+100 imes 40 ight]-91.32 imes 10^{-4} S_y[/latex]

i.e.
[latex]q_{\mathrm{b}, 23}=-10^{-4} S_y\left(0.12 s_3-0.15 imes 10^{-2} s_3^2+125.52 ight)[/latex] (iv)
The remaining qb distribution follows from symmetry. From Eq. (17.27)

[latex]q_{s, 0}=-\frac{\oint\left(q_{\mathrm{b}} / G t ight) \mathrm{d} s}{\oint \mathrm{d} s / G t}[/latex] (17.27)

[latex]q_{s, 0}=-\frac{\oint\left(q_{\mathrm{b}} / t ight) \mathrm{d} s}{\oint \mathrm{d} s / t}[/latex] (v)

in which

[latex]\oint \frac{\mathrm{d} s}{t}=\frac{80}{0.64}+\frac{2 imes 240}{0.64}+\frac{80}{0.36}=1097.2[/latex]

Now substituting in Eq. (v) for [latex]q_{\mathrm{b}, \mathrm{O} 1}, q_{\mathrm{b}, 12} ext { and } q_{\mathrm{b}, 23}[/latex] from Eqs (ii)–(iv), respectively

[latex]\begin{aligned}&q_{s, 0}=\frac{2 imes 10^{-4} S_y}{1097.2}\left[\int_0^{40} \frac{0.27 imes 10^{-2}}{0.64} s_1^2 \mathrm{~d} s_1+\int_0^{240} \frac{1}{0.64}\left(0.22 s_2+38.52 ight) \mathrm{d} s_2 ight.\&\left.+\int_0^{40} \frac{1}{0.64}\left(0.12 s_3-0.15 imes 10^{-2} s_3^2+125.52 ight) \mathrm{d} s_3 ight]\end{aligned}[/latex]

from which
[latex]q_{s, 0}=70.3 imes 10^{-4} S_y[/latex]
The complete shear flow distribution is then

[latex]q_{\mathrm{O} 1}=-10^{-4} S_y\left(0.27 imes 10^{-2} s_1^2-70.3 ight)[/latex] (vi)

[latex]q_{12}=q_{34}=-10^{-4} S_y\left(0.22 s_2-31.78 ight)[/latex] (vii)
[latex]q_{23}=-10^{-4} S_y\left(0.12 s_3-0.15 imes 10^{-2} s_3^2-55.22 ight)[/latex] (viii)
Taking moments about the mid-point of the wall 23

[latex]S_y \xi_{\mathrm{S}}=2\left[\int_0^{40} q_{01} imes 240 \mathrm{~d} s_1+\int_0^{240} q_{12} imes 40 \mathrm{~d} s_2 ight][/latex] (ix)

Substituting for [latex]q_{\mathrm{O} 1} ext { and } q_{12}[/latex] from Eqs (vi) and (vii) in Eq. (ix)

[latex]\begin{aligned}S_y \xi_{\mathrm{S}}=-2 imes 10^{-4} S_y & \left[\int_0^{40}\left(0.27 imes 10^{-2} s_1^2-70.3 ight) imes 240 \mathrm{~d} s_1 ight. \&\left.+\int_0^{240}\left(0.22 s_2-31.78 ight) imes 40 \mathrm{~d} s_2 ight]\end{aligned}[/latex]

from which
[latex]\xi_{\mathrm{S}}[/latex] = 142.5 mm

Question p.20.4: Find the position of the shear centre of the rectangular fou...

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